Question

In the following sequence of reactions,
$C{H}_3-Br\stackrel{KCN}{\to }A\stackrel{H_3{O}^{+}}{\to }B\underset{\text{ ether }}{\overset{LiAl{H}_4}{\to }}C$
the end product $C$ is

• A. Acetone
• B. Methane
• C. Acetaldehyde
• D. Ethyl alcohol

Answer 1

Answer: (D)

The end product $C$ is ethyl alcohol.
$C{H}_{3}Br\stackrel{KCN}{\to }\underset{A}{C{H}_3-CN}\stackrel{H_3{O}^{+}}{\to }\underset{B}{C{H}_3-COOH}\stackrel{LiAl{H}_4}{\to }\underset{C}{C{H}_3-C{H}_2-OH}$
Methyl bromide reacts with $KCN$ to form acetonitrile. $Br$ atom is replaced with $CN$ group.
Hydrolysis of cyano group gives carboxylic group. Reduction of carboxylic group gives hydroxyl group.